作业2 习题2

/*计算和，差，商，积*/
#include<stdio.h>
int main(void)
{
    
    double num1,num2;
    char op;
    printf("Type in an expression:");
    scanf("%lf%c%lf",&num1,&op,&num2);
    switch(op){
        case'+':
        printf("=%.2f\n",num1+num2);
        break;
        case'-':
        printf("=%.2f\n",num1-num2);
        break;
        case'*':
        printf("=%.2f\n",num1*num2);
        break;
        case'/':
        printf("=%.2f\n",num1/num2);
        break;        default:
        printf("SORRY,sorry\n");
        break;
    }
    return 0;
}

 

 

 

 

 

 

 

/*输入x,n，计算x^n*/
#include<stdio.h>
int main(void)
{
    int i,n; 
    double sum, x;
    printf("enter x,n:");
    scanf("%lf%d", &x, &n); 
    sum= 1; 
    for(i = 1; i <= n; i++) {
    sum= sum * x; 
    printf("%.2f\n", sum);
    }
     return 0; 
     } 

 

 

 

 

/*计算i^2+1/i的和*/
#include <stdio.h> 
int main(void)
{
    int i,m,n;
    double sum;
    printf("enter m and n:");
    scanf("%d%d",&m,&n);
    sum=0;      /*赋予sum的初值为0*/
    for(i=m;i<=n;i++){
        sum=sum+i*i+1.0/i;
    }
        printf("sum=%.2f\n",sum);
        return 0;
    } 

 

 

 

 

/*计算1-2/3+3/5-4/7+5/9-6/11+...的n项之和*/
#include<stdio.h>
#include<math.h>
int main(void)
{
    int i,n,fenzi,fenmu,flag;
    double item,sum;
    printf("enter n:");
    scanf("%d",&n);
    flag=1;         /*赋予初值*/
    fenzi=1;
    fenmu=1;
 sum=0;
    for(i=1;i<=n;i++){
        item=1.0*flag*fenzi/fenmu;
        
        sum=sum+item;
        fenzi=fenzi+1;
        fenmu=fenmu+2;
        flag=-flag;
    }
        printf("sum=%.2f\n",sum);
    
    return 0;
    
}

 

 

 

/*输出“还款年限-月还款表”*/
#include <stdio.h> 
#include <math.h>     /*程序中调用数学函数*/
int main(void) 
{  
    int year;  
    double loan,money,my,rate; 
    scanf("%Lf",&loan);
    scanf("%Lf",&rate);
    printf("enter year money\n"); 
    for(year=5;year<=30;year++){ 
        my=pow(1+rate,12*year);  
        money=loan*rate*my/(my-1); 
        printf("year=%d money=%.0f\n",year,money); 
    }  
    
    return 0; 
}  

 

 

 

/*调用pow函数求幂*/
#include<stdio.h>
#include<math.h>     /*调用数学函数*/
int main(void)
{
    int n,i;
    double power,sum;
    
    printf("enter n:");
    scanf("%d",&n);
    sum=0;
    for(i=1;i<=n;i++){
    power=pow(2,i)；      /*调用幂函数pow(2,i)计算2的i次方*/
    sum=sum+power;
    }
        printf("sum=%.2f\n",sum);
        return 0;
    }